In loop II (B-C-D-B), how are DC and BD treated if flows are D to C and B to D?

Both are taken as negative (against the clockwise direction)

2075 Bhadra · 8 marks

Hardy Cross with given constant factors

Short answer

After Hardy‑Cross iteration with resistance factors r (AB=1, BC=2, CD=1, DA=2, BD=3), the final discharges are: AB = 58.5 A→B, BC = 31.1 B→C, CD = 43.9 D→C, DA = 41.5 A→D, BD = 2.4 B→D, all in the same flow units.

Using Hardy-Cross method, find the rate of flow in every pipe lines as given below. The constant factor for AB, BC, CD, DA and BD are 1, 2, 1, 2 and 3 respectively. [Figure: network with nodes A, B, C, D and diagonal BD; 100 units enter at A; 25 units leave at B; 75 units leave at C]

Data. Constant factors ( $r$ in $h_f = rQ^2$ ): AB 1, BC 2, CD 1, DA 2, BD 3. 100 units enter at A; 25 leave at B and 75 at C.

Initial distribution (continuity): A to B 60, A to D 40; at B (out 25): B to C 20, B to D 15; at D: inflow $40 + 15 = 55$ all goes D to C; at C: $20 + 55 = 75$ .

Loop I = A-B-D-A (AB +, BD +, DA $-$ as AD). Loop II = B-C-D-B (BC +, CD $-$ as DC, DB $-$ as BD).

Hardy-Cross iterations

| Iter | $\sum rQ|Q|$ (I) | $2\sum r|Q|$ (I) | $\Delta Q_I$ | $\sum rQ|Q|$ (II) | $2\sum r|Q|$ (II) | $\Delta Q_{II}$ | |---|---|---|---|---|---|---| | 1 | 1075.00 | 370.00 | -2.905 | -2900.00 | 280.00 | 10.357 | | 2 | -412.90 | 296.24 | 1.394 | -158.93 | 221.14 | 0.719 | | 3 | -8.07 | 297.50 | 0.027 | -15.38 | 226.63 | 0.068 |

After 10 iterations the corrections are negligible.

Final discharges

Pipe	AB	BC	CD (D to C)	DA (A to D)	BD (B to D)
Q	58.52	31.15	43.85	41.48	2.37

Checks: A: $58.52 + 41.48 = 100.00$ ; B: $58.52 = 25 + 31.15 + 2.37$ ; C: $31.15 + 43.85 = 75.00$ ; D: $41.48 + 2.37 = 43.85$ .

Answer: AB = 58.5, BC = 31.1, CD = 43.9, DA = 41.5, and BD = 2.4 units (B to D).

✓ Answer checked by Claude Opus 5.5 (high effort)

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